From previous episodes of Amnesia's Dream,
\(\cfrac { \partial T }{ \partial t } =\cfrac { x }{ g_{ T } } \cfrac { \partial ^{ 3 }T }{ \partial ^{ 3 }t } \)
and so,
\(\cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \cfrac { \partial }{ \partial t } (\cfrac { \partial ^{ 2 }T }{ \partial t^{ 2 } } )\)
From the wave equation,
\( \cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \cfrac { \partial }{ \partial t } (v^{ 2 }\cfrac { \partial ^{ 2 }T }{ \partial x^{ 2 } } )\)
\( \cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \left\{ 2v\cfrac { dv }{ dt } \cfrac { \partial ^{ 2 }T }{ \partial x^{ 2 } } +v^{ 2 }\cfrac { \partial ^{ 2 } }{ \partial x^{ 2 } } (\cfrac { \partial T }{ \partial t } ) \right\} \)
If we assume that acceleration is wholly due to \(g_T\),
\(\cfrac{dv}{dt}=g_T\)
\( \cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \left\{ 2vg_{ T }T(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } +v^{ 2 }\cfrac { \partial ^{ 2 } }{ \partial x^{ 2 } } (T(x)\cfrac { dT(t) }{ dt } ) \right\} \)
\( \cfrac { dT(t) }{ dt } T(x)=\cfrac { x }{ g_{ T } } \left\{ 2vg_{ T }T(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } +v^{ 2 }\cfrac { dT(t) }{ dt } \cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } \right\} \)
\( \cfrac { dT(t) }{ dt } T(x)=2xvT(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } +\cfrac { dT(t) }{ dt } v^{ 2 }\cfrac { x }{ g_{ T } } \cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } \)
Compare this with,
\(T(x)= v^{ 2 }\cfrac { x }{ g_{ T } } \cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } \)
this suggests,
\( 2xvT(t)\cfrac { d^{ 2 }T(x) }{ dx^{ 2 } } =2xv\cfrac { \partial ^{ 2 }T }{ \partial x^{ 2 } } =\cfrac { 2x }{ v } \cfrac { \partial ^{ 2 }T }{ \partial t^{ 2 } } =0\)
which might mean,
\(\cfrac { \partial^{ 2 }T }{ \partial x^{ 2 } } =\cfrac { \partial ^{ 2 }T }{ \partial t^{ 2 } }=0\)
then there is no wave in the radial direction, not of the form \(T=T(t)T(x)\). Just attenuation.
